Let $a$ and $b$ be angles such that
\[\cos (a + b) = \cos a + \cos b.\]Find the maximum value of $\cos a.$
From $\cos (a + b) = \cos a + \cos b,$ $\cos a = \cos (a + b) - \cos b.$  Then from sum-to-product,
\[\cos (a + b) - \cos b = -2 \sin \frac{a + 2b}{2} \sin \frac{a}{2}.\]Let $k = \sin \frac{a + 2b}{2},$ so
\[\cos a = -2k \sin \frac{a}{2}.\]Then
\[\cos^2 a = 4k^2 \sin^2 \frac{a}{2} = 4k^2 \cdot \frac{1}{2} (1 - \cos a) = 2k^2 (1 - \cos a),\]so
\[\frac{\cos^2 a}{1 - \cos a} = 2k^2 \le 2.\]Then $\cos^2 a \le 2 - 2 \cos a,$ so
\[\cos^2 a + 2 \cos a + 1 \le 3.\]This means $(\cos a + 1)^2 \le 3,$ so $\cos a + 1 \le \sqrt{3},$ or $\cos a \le \sqrt{3} - 1.$

Equality occurs if we take $a = \arccos (\sqrt{3} - 1)$ and $b = \frac{3 \pi - a}{2}$ (which will make $k = \sin \frac{a + 2b}{2} = -1$), so the maximum value of $\cos a$ is $\boxed{\sqrt{3} - 1}.$